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We have a continuous and increasing function $f:[0,1]\to \mathbb R$ and the sequence $(a_n)_{n\ge1}$,$$a_n=\frac{1}{2^n}\sum_{k=1}^{2^n}f\biggl(\frac{k}{2^n}\biggl)$$ . Knowing that there exists $p\in \mathbb N^*$ s.t. $a_p=\int_0^1f(x)dx$, prove that f is constant.

I've found that $\int_0^1f(x)dx=a_p\ge a_{p+n}\ge \int_0^1f(x)dx $ because $a_n$ is decreasing (proof here: $a_n=\frac{1}{2^n}\sum_{k=1}^{2^n}f\biggl(\frac{k}{2^n}\biggl)$ is decreasing ) and since the integral is the limit of $a_{p+n}$, the limit has to be smaller than it.

So $a_{p+n}=\int_0^1f(x)dx ,\forall n\in \mathbb N$.

I don't know what can I do next.Can somebody help me,please?

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    $\begingroup$ To get an idea consider a simple example like $p=1$. You are given that $f(1/2) + f(1) = 2\int_0^1 f(x){\rm d}x$ and that $f$ is increasing and continuous. Split the integral in two: $2\int_0^{1/2}f(x){\rm d}x + 2\int_{1/2}^1f(x){\rm d}x$. Notice that the first integral has to be less than $f(1/2)$ and the second one is less than $f(1)$ if $f$ is not constant. This gives a contradiction. $\endgroup$ – Winther Mar 5 at 12:10
  • $\begingroup$ @Gaboru: You have clearly put some thought into the problem. If you can, for problems like this, please also include the source where you encountered the problem. You can find more advice on composing posts at "How to ask a good question", math.meta.stackexchange.com/questions/9959/… $\endgroup$ – Carl Mummert Mar 5 at 12:12
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Hints: let $g(x)=f(\frac {k-1} {2^{p}})$ if $\frac {k-1} {2^{p}} \leq x <\frac k {2^{p}}$. Verify that $f \geq g$ and $\int_0^{1} [f(x)-g(x)]\, dx=0$. Conclude that $f$ is a constant in each of the intervals $\frac {k-1} {2^{p}} \leq x <\frac k {2^{p}}$, hence on $[0,1]$.

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  • $\begingroup$ Can you give me more details, please? $\endgroup$ – Gaboru Mar 15 at 18:59
  • $\begingroup$ @Gaboru $f \geq g$ on $[\frac {k-1} {2^{p}},\frac k {2^{p}})$ because $f(x) \geq f(\frac {k-1} {2^{p}})$. The equation $\int (f-g)=0$ follows from the hypothesis if you split the integral into integrals over the intervals $[\frac {k-1} {2^{p}},\frac k {2^{p}})$. If the integral of a non-negative function is $0$ then the fuction is $0$ almost everywhere and since $f$ is continuous we see that it is a constant on each of the intervals $[\frac {k-1} {2^{p}},\frac k {2^{p}})$. $\endgroup$ – Kavi Rama Murthy Mar 15 at 23:05
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We need the following lemma: if $g:[0,1]\to \Bbb R$ is a continuous, increasing function such that $$ \int_0^1 g(t)\ dt= g(1), $$ then $g(x)=g(1)$ for all $x\in [0,1]$. Proof is simple. Since $g(1)-g(x)\ge 0$ is a continuous function with $ \int_0^1 g(1)-g(t)\ dt=0 $, it follows that $g(1)-g(x)$ is identically $0$.

Now, define $g(x) = \frac1{2^n}\sum_{k=0}^{2^n-1}f(\frac{k+x}{2^n})$. Then the given condition implies that $$\begin{align*} \int_0^1 g(t)\ dt &= \sum_{k=0}^{2^n-1}\frac1{2^n}\int_0^1f\left(\frac{k+t}{2^n}\right)\ dt\\&=\sum_{k=0}^{2^n-1}\int_{\frac k{2^n}}^{\frac{k+1}{2^n}}f\left(u\right)\ du\\&=\int_0^1 f(u)\ dt\\&=\frac1{2^n}\sum_{k=1}^{2^n} f\left(\frac{k}{2^n}\right)=g(1). \end{align*}$$ Since $g$ is increasing, the lemma in particular implies that $g(0)=g(1)$, hence $$ \sum_{k=0}^{2^n-1} f\left(\frac{k}{2^n}\right)=\sum_{k=1}^{2^n} f\left(\frac{k}{2^n}\right)\implies f(0)=f(1). $$ Since $f$ is increasing, this implies that $f$ is constant.

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